Journal of Algebraic Combinatorics 12 (2000), 25–36
c 2000 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
Distance-Regular Graphs Related to the Quantum
Enveloping Algebra of sl(2)
BRIAN CURTIN
Department of Mathematics, University of California, Berkeley CA 94720, USA
curtin@math.berkeley.edu
KAZUMASA NOMURA
nomura@tmd.ac.jp
College of Liberal Arts and Sciences, Tokyo Medical and Dental University, Kohnodai, Ichikawa, 272 Japan
Received May 5, 1998; Revised April 21, 1999
Abstract. We investigate a connection between distance-regular graphs and Uq (sl(2)), the quantum universal
enveloping algebra of the Lie algebra sl(2). Let 0 be a distance-regular graph with diameter d ≥ 3 and valency
k ≥ 3, and assume 0 is not isomorphic to the d-cube. Fix a vertex x of 0, and let T = T (x) denote the Terwilliger
algebra of 0 with respect to x. Fix any complex number q 6∈ {0, 1, −1}. Then T is generated by certain matrices
satisfying the defining relations of Uq (sl(2)) if and only if 0 is bipartite and 2-homogeneous.
Keywords: distance-regular graph, Terwilliger algebra, quantum group
1.
Introduction
We investigate a connection between distance-regular graphs and Uq (sl(2)), the quantum
universal enveloping algebra of the Lie algebra sl(2). It is well-known that there is a
“natural” sl(2) action on the d-cubes (see Proctor [9] or Go [4]). Here we describe the
distance-regular graphs with a similar natural Uq (sl(2)) action. We show that these graphs
are precisely the bipartite distance-regular graphs which are 2-homogeneous in the sense
of [7, 8], excluding the d-cubes. To state this precisely, we recall some definitions.
Let U (sl(2)) denote the unital associative C-algebra generated by X − , X + , and Z subject
to the relations
Z X − − X − Z = 2X − ,
Z X + − X + Z = −2X + ,
X−X+ − X+X− = Z.
(1)
U (sl(2)) is called the universal enveloping algebra of sl(2). For any complex number q
satisfying
q 6= 1,
q 6= 0,
q 6= −1,
(2)
−
+
let Uq (sl(2)) denote the unital associative C-algebra generated by X , X , Y , and Y −1
subject to the relations
Y Y −1 = Y −1 Y = 1,
Y X − = q 2 X − Y,
Y X + = q −2 X + Y,
(3)
X−X+ − X+X− =
−1
Y −Y
.
q − q −1
(4)
26
CURTIN AND NOMURA
Uq (sl(2)) is called the quantum universal enveloping algebra of sl(2). For more on Uq (sl(2))
and its relation to U (sl(2)) see [5, 6].
Let 0 = (X, R) denote a finite, undirected, connected graph without loops or multiple
edges and having vertex set X , edge set R, distance function ∂, and diameter d. 0 is said to
be distance-regular whenever for all integers `, i, j (0 ≤ `, i, j ≤ d) there exists a scalar
pi`j such that for all x, y ∈ X with ∂(x, y) = `, |{z ∈ X | ∂(x, z) = i, ∂(y, z) = j}| = pi`j .
i
i
Assume that 0 is distance-regular. Set c0 = 0, ci = p1i−1
(1 ≤ i ≤ d), ai = p1i
i
(0 ≤ i ≤ d), bi = p1i+1 (0 ≤ i ≤ d − 1), and bd = 0. 0 is regular with valency
0
, and ci + ai + bi = k (0 ≤ i ≤ d). 0 is bipartite precisely when ai = 0
k = b0 = p11
(0 ≤ i ≤ d).
Let 0 = (X, R) denote a bipartite distance-regular graph. 0 is said to be 2-homogeneous
whenever for all integers i (1 ≤ i ≤ d) there exists a scalar γi such that for all x, y,
z ∈ X with ∂(x, y) = i, ∂(x, z) = i, ∂(y, z) = 2, |{w ∈ X | ∂(x, w) = i − 1, ∂(y, w) =
1, ∂(z, w) = 1}| = γi . 0 may be 2-homogeneous despite the fact that some structure
constant γi is not uniquely determined: This occurs when there are no x, y, z ∈ X with
∂(x, y) = i, ∂(x, z) = i, ∂(y, z) = 2. It is known that γd is not uniquely determined
when 0 is 2-homogeneous [8]. The d-cube is the graph with vertex set X = {0, 1}d (the
d-tuples with entries in {0, 1}) such that two vertices are adjacent if and only if they differ in
precisely one coordinate. The d-cube is a 2-homogeneous bipartite distance-regular graph
with γi = 1 (1 ≤ i ≤ d − 1). The 2-homogeneous bipartite distance-regular graphs have
been studied in [3, 8, 11].
Let Mat X (C) denote the C-algebra of matrices with rows and columns indexed by X . Let
A ∈ Mat X (C) denote the adjacency matrix of 0. For the rest of this section fix x ∈ X . For
all i (0 ≤ i ≤ d), define E i∗ = E i∗ (x) to be the diagonal matrix in Mat X (C) such that for
all y ∈ X , E i∗ has (y, y)-entry equal to 1 if ∂(x, y) = i, and 0 otherwise. Let T = T (x)
denote the subalgebra of Mat X (C) generated by A, E 0∗ , E 1∗ , . . . , E d∗ .
Pd
Pd−1 ∗
∗
∗
E i AE i+1
and R = i=1
E i∗ AE i−1
. Proctor [9] showed that if 0 is isoSet L = i=0
Pd
−
morphic to the d-cube, then the matrices X = L, X + = R, and Z = i=0
(d − 2i)E i∗
satisfy the relations of (1) (see also Go [4]). We must slightly relax the form of these
matrices to admit a Uq (sl(2)) structure. Specifically, we consider matrices of the form:
X− =
d−1
X
i=0
∗
xi− E i∗ AE i+1
,
X+ =
d
X
i=1
∗
xi+ E i∗ AE i−1
,
Y =
d
X
yi E i∗ ,
(5)
i=0
where xi− (0 ≤ i ≤ d − 1), xi+ (1 ≤ i ≤ d), and yi (0 ≤ i ≤ d) are arbitrary complex
Pd
yi−1 E i∗ .
scalars. Y is invertible if and only if yi 6= 0 (0 ≤ i ≤ d), in which case Y −1 = i=0
Theorem 1.1 Let 0 = (X, R) denote a distance-regular graph with diameter d ≥ 3 and
valency k ≥ 3. Assume that 0 is not isomorphic to the d-cube. Fix x ∈ X , and write
E i∗ = E i∗ (x) (0 ≤ i ≤ d) and T = T (x). Let X − , X + , and Y be any matrices of the form
(5), and let q be any nonzero complex number. Then the following are equivalent.
(i) Y is invertible, X − , X + , Y , Y −1 generate T , and (2)–(4) hold.
(ii) 0 is bipartite and 2-homogeneous, (q + q −1 )2 = c22 b2−1 (k − 2)(c2 − 1)−1 , and there
27
QUANTUM ENVELOPING ALGEBRA
exists ² ∈ {1, −1} such that
yi
− +
xi xi+1
= ²q d−2i
= ²q
−2i+1
(0 ≤ i ≤ d),
(q d + q 2i )(q d + q 2i+2 )(q d + q 2 )−2
(0 ≤ i ≤ d − 1).
The condition (i) of Theorem 1.1 means that the Terwilliger algebra T is a homomorphic
image of Uq (sl(2)). The factor of ² appears in (ii) because the defining relations of Uq (sl(2))
are invariant under changing the signs of any two of X − , X + , and Y .
2.
Background
Throughout this section, let 0 = (X, R) denote a distance-regular graph with diameter d.
Let Mat X (C) denote the C-algebra of matrices with rows and columns indexed by X . For
all i (0 ≤ i ≤ d), define Ai to be the matrix in Mat X (C) such that for all y, z ∈ X the
Observe that A0 = I (the identity
(y, z)-entry of Ai is 1 if ∂(y, z) = i and 0 otherwise. P
d
matrix
of
0,
and
matrix), A := A1 is the adjacency
i=0 Ai = J (the all 1’s matrix).
Pd
Observe that Ai A j = A j Ai = `=0 pij` A` (0 ≤ i, j ≤ d). It follows that the linear span
M of A0 , A1 , . . . , Ad is a commutative subalgebra of Mat X (C). The algebra M is called
the Bose-Mesner algebra of 0. It is known that M is generated by A. See [1, 2] for more
on distance-regular graphs and their Bose-Mesner algebras.
For the rest of this section fix x ∈ X . For all i (0 ≤ i ≤ d), define E i∗ = E i∗ (x)
of E i∗
to be the diagonal matrix in Mat X (C) such that for all y ∈ X , the (y, y)-entry
Pd
is E i∗ (y, y) = Ai (x, y). Observe that E i∗ E ∗j = δij E i∗ (0 ≤ i, j ≤ d) and i=0
E i∗ = I .
It follows that the linear span M∗ = M∗ (x) of E 0∗ , E 1∗ , . . . , E d∗ is a commutative subalgebra
of Mat X (C). The algebra M∗ is called the dual Bose-Mesner algebra of 0 with respect to x.
Let T = T (x) denote the subalgebra of Mat X (C) generated by M ∪ M∗ . The algebra T
is called the Terwilliger algebra of 0 with respect to x. See [10] for more on Terwilliger
algebras.
Fix `, i, j (0 ≤ `, i, j ≤ d). Observe that for all y, z ∈ X , the (y, z)-entry of E i∗ A` E ∗j
is 0 or 1, and it is equal to 1 if and only if ∂(x, y) = i, ∂(y, z) = ` and ∂(x, z) = j. Thus,
considering the positions of the nonzero entries,
{E i∗ A` E ∗j 6= 0 | 0 ≤ `, i, j ≤ d} is linearly independent,
E i∗ A` E ∗j
6= 0 if and only if
pi`j
6= 0.
(6)
(7)
6 0
Observe that pi`j = 0 if one of `, i, j is greater than the sum of the other two, and pij` =
∗
∗
∗
∗
AE
=
E
AE
=0
if one of `, i, j is equal to the sum of
the
other
two.
It
follows
that
E
i
j
j
i
Pd Pd
∗
∗
E
AE
=
L
+
F
+
R,
where
whenever |i − j| > 1. Hence A = i=0
j
j=0 i
L=
d−1
X
∗
E i∗ AE i+1
,
i=0
E i∗ AE i∗
F=
d
X
i=0
E i∗ AE i∗ ,
R=
d
X
∗
E i∗ AE i−1
.
i=1
= 0 if and only if ai = 0, so 0 is bipartite if and only if F = 0.
Observe that
We wish to emphasize the following combinatorial interpretation of L and R. For all
i (0 ≤ i ≤ d) and for all y ∈ X , let 0i (y) = {z ∈ X | ∂(y, z) = i}. Identify each
28
CURTIN AND NOMURA
vertex with its characteristic column vector, and note that Mat X (C) actsP
on the vertices by
(x),
L
y
=
left multiplication.
For
all
i
(0
≤
i
≤
d)
and
all
y
∈
0
i
w∈01 (y)∩0i−1 (x) w,
P
Ry = w∈01 (y)∩0i+1 (x) w, and E ∗j y = δij y (0 ≤ j ≤ d). Fix i (0 ≤ i ≤ d). For all y,
z ∈ 0i (x), set
β(y, z) = |01 (y) ∩ 01 (z) ∩ 0i+1 (x)|,
γ (y, z) = |01 (y) ∩ 01 (z) ∩ 0i−1 (x)|.
(8)
Observe that for all y, z ∈ 0i (x),
(LREi∗ )(y, z) = β(y, z),
(RLEi∗ )(y, z) = γ (y, z).
(9)
In particular, (LREi∗ )(y, y) = bi , (RLEi∗ )(y, y) = ci , and when ∂(y, z) > 2, (LREi∗ )(y, z) =
(RLEi∗ )(y, z) = 0.
3.
Construction of U(sl(2)) and Uq (sl(2)) structures
In this section, we construct a U (sl(2)) structure on the d-cubes and a Uq (sl(2)) structure on
the remaining 2-homogeneous bipartite distance-regular graphs. Throughout this section,
let 0 = (X, R) denote a distance-regular graph with diameter d ≥ 3 and valency k ≥ 3.
Fix x ∈ X , and write E i∗ = E i∗ (x) (0 ≤ i ≤ d), M∗ = M∗ (x), T = T (x).
Lemma 3.1 Let z 0 , z 1 , . . . , z d denote distinct complex scalars. Then Z =
generates M∗ .
Pd
∗
i=0 z i E i
Pd
j ∗
j
Proof: Observe that
where the j = 0 equation is
PdZ =∗ i=0 z i E i ∗(0 ≤∗ j ≤ d),
∗
E
.
Viewing
E
,
E
,
.
.
.
,
E
as
unknowns, this is a system
interpreted as I =
d
i=0 i
0
1
of linear equations with Vandermonde (hence invertible) coefficient matrix. Thus E i∗ ∈
span {Z j | 0 ≤ j ≤ d} (0 ≤ i ≤ d), so Z generates M∗ .
2
Lemma
0 is isomorphic to the d-cube. Then X − = L, X + = R and
Pd 3.2 [4, 9] Suppose
∗
Z = i=0 (d − 2i)E i generate T and satisfy (1).
Proof: Observe that Z generates M∗ by Lemma 3.1. Observe that F = 0 since 0 is
bipartite, so A = L + R. A generates M, so L, R, and Z generate T .
The relations Z L − L Z = 2L and Z R − R Z = −2R are easily verified using the
definitions of L, R, and Z and
that E i∗ E ∗j = δij E i∗ (0 ≤ i, j ≤ d). It remains to
Pdthe fact
∗
verify LR − RL = Z . Since i=0 E i = I , it is enough to show that for all i (0 ≤ i ≤ d)
LREi∗ − RLEi∗ = (d − 2i)E i∗ .
(10)
Fix i (0 ≤ i ≤ d), and pick y, z ∈ 0i (x). Let r , s, t denote the (y, z)-entries of LREi∗ ,
RLEi∗ , and E i∗ , respectively. From (8), (9) we find the following. Suppose ∂(y, z) > 2.
Then r = s = t = 0. Suppose ∂(y, z) = 2. Then r = c2 − γi = 1, s = γi = 1, and
t = 0. The case ∂(y, z) = 1 does not occur since ai = 0. Finally suppose y = z. Then
r = bi = d − i, s = ci = i, and t = 1. In all cases r − s = (d − 2i)t, so (10) holds. 2
29
QUANTUM ENVELOPING ALGEBRA
Theorem 3.3 ([3, Theorem 35]) Suppose 0 is not isomorphic to the d-cube. Then 0 is
bipartite and 2-homogeneous if and only if there exists a complex scalar q 6∈ {0, 1, −1}
such that
ci = ei [ i ],
bi = ei [ d − i ] (0 ≤ i ≤ d),
(11)
where
ei = q i−1 (q d + q 2 )(q d + q 2i )−1 ,
[ i ] = (q i − q −i )(q − q −1 )−1
(12)
for all integers i. Suppose the above equivalent conditions hold. Then
−1
γi = e2 ei ei+1
(1 ≤ i ≤ d − 1).
(13)
Corollary 3.4 ([3, Corollary 36]) Suppose 0 is bipartite and 2-homogeneous, but not
isomorphic to the d-cube. Then any complex scalar q 6∈ {0, 1, −1} satisfying (11) and (12)
is real and
(q + q −1 )2 = c22 b2−1 (b0 − 2)(c2 − 1)−1 .
(14)
The set of q satisfying (14) is of the form {λ, λ−1 , −λ, −λ−1 } for some real number
λ > 1. When d is even, all such q satisfy (11). When d is odd, only q ∈ {λ, λ−1 } satisfy (11)
since q + q −1 = c2 γr−1 > 0, where r = (d − 1)/2 (see [3, Corollary 36]).
Lemma 3.5 Suppose 0 is bipartite and 2-homogeneous but not isomorphic to the d-cube.
Let q 6∈ {0, 1, −1} be any complex scalar such that (11), (12) hold, and let ei (0 ≤ i ≤ d)
be as in (12). Then the matrices
X− =
d−1
X
∗
∗
e−1
j E j AE j+1 ,
j=0
X+ =
d
X
∗
∗
e−1
j E j AE j−1 ,
j=1
Y =
d
X
q d−2 j E ∗j
j=0
generate T and satisfy (4).
Pd−1
Proof:P Observe that Y generates M∗ by Lemma 3.1. Now L = ( i=0
ei E i∗ )X − and
d
∗
+
−
+
R = ( i=1 ei E i )X are in the algebra generated by Y , X and X . Observe that F = 0
since 0 is bipartite, so A = L + R. A generates M, so X − , X + , and Y generate T .
The relations Y X − = q 2 X − Y and Y X + = q −2 X + Y are easily verified using the definitions of X − , X + , and Y and the fact that E i∗ E ∗j = δij E i∗ (0 ≤ i, j ≤ d). It remains to verify
X − X + − X + X − = (Y − Y −1 )/(q − q −1 ). Observe that for all i (0 ≤ i ≤ d), X − X + E i∗ =
−1
−1 −1
LREi∗ , X + X − E i∗ = ei−1
ei RLEi∗ , and (Y − Y −1 )/(q − q −1 )E i∗ = [ d − 2i ]E i∗ .
ei−1 ei+1
Pd
∗
Thus, since I = i=0 E i , it is enough to show that for all i (0 ≤ i ≤ d)
−1
−1 −1
LREi∗ − ei−1
ei RLEi∗ = [ d − 2i ]E i∗ .
ei−1 ei+1
(15)
30
CURTIN AND NOMURA
Fix i (0 ≤ i ≤ d), and pick y, z ∈ 0i (x). Let r , s, and t denote the (y, z)-entries of LREi∗ ,
RLEi∗ , and E i∗ , respectively. From (8), (9) we find the following. Suppose ∂(y, z) > 2.
Then r = s = t = 0. Suppose ∂(y, z) = 2. Then by the definition of 2-homogeneous,
r = c2 − γi , s = γi , and t = 0. It can be verified by a direct computation using (11)–(13)
−1
−1 −1
(c2 − γi ) − ei−1
ei γi = 0. The case ∂(y, z) = 1 does not occur since ai = 0.
that ei−1 ei+1
Finally suppose y = z. Then r = bi , s = ci , and t = 1. It can be verified by a direct
−1
−1 −1
bi − ei−1
ei ci = [ d − 2i ]. In all
(albeit long) computation using (11), (12) that ei−1 ei+1
−1 −1
−1 −1
2
cases ei ei+1r − ei−1 ei s = [ d − 2i ]t, so (15) holds.
The U (sl(2)) structure on the d-cube is very similar to the Uq (sl(2)) structure on the
remaining 2-homogeneous bipartite distance-regular graphs. In the sequel, we exploit this
similarity to prove the following result and Theorem 1.1 simultaneously.
Theorem 3.6 Let 0 = (X, R) denote a distance-regular graph with diameter d ≥ 3 and
valency k ≥ 3. Fix x ∈ X , and write E i∗ = E i∗ (x) (0 ≤ i ≤ d), T = T (x). Let X − , X + ,
Pd−1 − ∗
Pd
Pd
∗
∗
xi E i AE i+1
, X + = i=1
xi+ E i∗ AE i−1
, Z = i=0
z i E i∗
and Z be of the form X − = i=0
−
+
for some complex scalars xi (0 ≤ i ≤ d − 1), xi (1 ≤ i ≤ d), z i (0 ≤ i ≤ d). Then the
following are equivalent.
(i) X − , X + , and Z generate T and satisfy (1).
(ii) 0 is isomorphic to the d-cube, and
+
xi− xi+1
=1
(0 ≤ i ≤ d − 1),
z i = d − 2i
(0 ≤ i ≤ d).
As in Theorem 1.1, The condition (i) of Theorem 3.6 means that the Terwilliger algebra T
is a homomorphic image of U (sl(2)).
4.
Combinatorial structure
We show that the U (sl(2)) and Uq (sl(2)) structures of Lemmas 3.2 and 3.5 can only occur
on a 2-homogeneous bipartite distance-regular graph. Specifically, we show the following.
Theorem 4.1 Let 0 = (X, R) denote a distance-regular graph with diameter d ≥ 3 and
valency k ≥ 3. Fix x ∈ X, and write E i∗ = E i∗ (x) (0 ≤ i ≤ d), T = T (x). Suppose
X − X + − X + X − = Z , where X − and
that T is generated by {X − , X + } ∪ M∗ and that P
d
+
z i E i∗ for some complex scalars z i
X are of the form (5) and Z is of the form Z = i=0
(0 ≤ i ≤ d). Then 0 is bipartite and 2-homogeneous.
The hypotheses of this result are met by both Theorems 1.1(i) and 3.6(i). Throughout
this section, we adopt the notation and assumptions of Theorem 4.1 as we prove this result
in a series of lemmas. The first step in our proof of Theorem 4.1 is to show that ai = 0
(1 ≤ i ≤ d − 1). To do so, we consider certain matrices in the left ideal T E 1∗ of T :
K i = E i∗ JE∗1
N0 = 0,
(0 ≤ i ≤ d),
Ni = E i∗ Ai−1 E 1∗
(1 ≤ i ≤ d).
31
QUANTUM ENVELOPING ALGEBRA
Lemma 4.2 LK 0 = 0, LK i = bi−1 K i−1 (1 ≤ i ≤ d), RK i = ci+1 K i+1 (0 ≤ i ≤ d − 1),
−
+
RK d = 0, and X − K 0 = 0, X − K i = xi−1
bi−1 K i−1 (1 ≤ i ≤ d), X + K i = xi+1
ci+1 K i+1
(0 ≤ i ≤ d − 1), X + K d = 0.
Proof: Clearly LK 0 = LE∗0 K 0 = 0. Fix i (1 ≤ i ≤ d). Fix y, z ∈ X , and let r and s
denote the (y, z)-entries of LK i and K i−1 , respectively. Observe that r = s = 0 unless
y ∈ 0i−1 (x) and z ∈ 01 (x), so suppose y ∈ 0i−1 (x) and z ∈ 01 (x). Then
∗
r = (E i−1
AEi∗ JE∗1 )(y, z) =
=
s=
X
X
∗
E i−1
(y, y)A(y, p)E i∗ ( p, p)J ( p, z)E 1∗ (z, z)
p∈X
A(y,
p)E i∗ ( p,
p∈X
∗
JE∗1 )(y, z)
(E i−1
p) = |01 (y) ∩ 0i (x)| = bi−1 ,
∗
= E i−1
(y, y)J (y, z)E 1∗ (z, z) = 1.
In all cases r = bi−1 s, so LK i = bi−1 K i−1 . The equations for RK i are proved similarly.
−
LEi∗ (1 ≤ i ≤ d) and
The equations involving X − and X + follow since X − E i∗ = xi−1
+
∗
∗
2
X + E i = xi+1 REi (0 ≤ i ≤ d − 1).
+
+
Lemma 4.3 xi− 6= 0 and xi+1
6= 0 (0 ≤ i ≤ d − 1). In particular, si := xi− xi+1
6= 0
(0 ≤ i ≤ d − 1).
Proof: Suppose xi− = 0 for some i (0 ≤ i ≤ d − 1), and set U = span{K h | i + 1 ≤ h
≤ d}. Then U is closed under left multiplication by the generators X − , X + , and M∗ of T
by Lemma 4.2 and construction. Hence U is a left ideal of T . However, LK i+1 = bi K i 6= 0
and K i 6∈ U, a contradiction. Hence xi− 6= 0 (0 ≤ i ≤ d − 1). A similar argument shows
+
6= 0 (0 ≤ i ≤ d − 1).
2
that xi+1
Lemma 4.4
+
X + Ni = xi+1
ci Ni+1 (1 ≤ i ≤ d − 1) and X + Nd = 0.
Proof: Fix i (1 ≤ i ≤ d − 1). Pick y, z ∈ X , and let r and s denote the (y, z)-entries of
X + Ni and Ni+1 , respectively. Observe that r = s = 0 unless y ∈ 0i+1 (x) and z ∈ 01 (x),
so suppose y ∈ 0i+1 (x) and z ∈ 01 (x). Then
+
∗
r = xi+1
(E i+1
AE i∗ Ai−1 E 1∗ )(y, z)
X
+
∗
E i+1
(y, y)A(y, p)E i∗ ( p, p)Ai−1 ( p, z)E 1∗ (z, z)
= xi+1
p∈X
=
s=
+
xi+1
|01 (y) ∩ 0i (x) ∩ 0i−1 (z)|,
∗
(E i+1 Ai E 1∗ )(y, z) = Ai (y, z).
+
Observe that r = s = 0 when ∂(y, z) 6= i, and r = xi+1
ci , s = 1 when ∂(y, z) = i. In all
+
+
+
2
cases r = xi+1 ci s, so X Ni = xi+1 ci Ni+1 . Clearly X + Nd = X + E d∗ Nd = 0.
Lemma 4.5
X − Ni ∈ span{Ni−1 , K i−1 } (1 ≤ i ≤ d).
32
CURTIN AND NOMURA
Proof: It is easy to show that X − N1 = x0− K 0 by entry-wise computation. We proceed
by induction: Fix i (2 ≤ i ≤ d), and assume X − Ni−1 = g Ni−2 + h K i−2 for some scalars
g, h. We compute
X − X + Ni−1 = X − (xi+ ci−1 Ni ) = xi+ ci−1 (X − Ni ),
+
+
ci−2 Ni−1 + hxi−1
ci−1 K i−1 ,
X + X − Ni−1 = X + (g Ni−2 + h K i−2 ) = gxi−1
Z Ni−1 = z i−1 Ni−1 .
Now we may apply the relation X − X + − X + X − = Z to Ni−1 and solve to find X − Ni ∈
2
span{Ni−1 , K i−1 } since xi+ ci−1 6= 0. The result follows by induction.
Lemma 4.6 ai = 0 (1 ≤ i ≤ d − 1).
Proof: By Lemmas 4.2–4.5 and construction, U = span{K i | 0 ≤ i ≤ d} + span{Ni | 1 ≤
i ≤ d} is a left ideal of T . In fact, U = T E 1∗ since E 1∗ = N1 . Now fix i (1 ≤ i ≤
d − 1). Then E i∗ T E 1∗ = E i∗ U = span{E i∗ K i , E i∗ Ni }, so dimC E i∗ T E 1∗ ≤ 2. Observe
that the subspace E i∗ T E 1∗ contains E i∗ A j E 1∗ ( j = i − 1, i, i + 1), and E i∗ Ai−1 E 1∗ 6= 0,
E i∗ Ai+1 E 1∗ 6= 0 by (7). If E i∗ Ai E 1∗ 6= 0, then these three matrices are linearly independent
2
by (6), contradicting dimC E i∗ T E 1∗ ≤ 2. Thus E i∗ Ai E 1∗ = 0, so ai = 0 by (7).
We show that ad = 0 by showing that there is a unique vertex at distance d from x.
+
(0 ≤ i ≤ d − 1) and s−1 = sd = 0. Then for all i
Lemma 4.7 Set si = xi− xi+1
(0 ≤ i ≤ d),
si LREi∗ − si−1 RLEi∗ = z i E i∗ ,
si β(y, z) − si−1 γ (y, z) = δ yz z i
(y, z ∈ 0i (x)),
(16)
(17)
where β(y, z) and γ (y, z) are as in (8). In particular, β(y, z) = 0 if and only if γ (y, z) = 0
for any distinct y, z ∈ 0i (x).
Proof: Fix i (0 ≤ i ≤ d). Apply the relation X − X + − X + X − = Z to E i∗ to get (16).
Fix y, z ∈ 0i (x). Computing the (y, z)-entry of (16) gives (17) by (9). It is clear from (17)
and Lemma 4.3 that β(y, z) = 0 if and only if γ (y, z) = 0 when y, z are distinct.
2
Lemma 4.8 |0d (x)| = 1 and 0 is bipartite.
Proof: By a down-up walk of length 2` (1 ≤ ` ≤ d), we mean a sequence of vertices
v0 , v1 , . . . , v2` such that vi and vi+1 are adjacent (0 ≤ i ≤ 2` − 1), vi , v2`−i ∈ 0d−i (x)
(0 ≤ i ≤ `), and v0 6= v2` . Assume |0d (x)| ≥ 2. For all distinct y, z ∈ 0d (x) there exists
a down-up walk of length 2d (taking v0 = y, vd = x, v2d = z), but there is no down-up
walk of length 2 since |0d−1 (x) ∩ 01 (y) ∩ 01 (z)| = 0 by Lemma 4.7.
Fix a down-up walk v0 , v1 , . . . , v2` of minimal length 2`. By minimality of the length of
this down-up walk, v`−1 and v`+1 ∈ 0d−`+1 (x) are distinct. Let γ (v`−1 , v`+1 ), β(v`−1 , v`+1 )
QUANTUM ENVELOPING ALGEBRA
33
be as in (8). Observe that γ (v`−1 , v`+1 ) > 0, so β(v`−1 , v`+1 ) > 0 by Lemma 4.7. Fix
w ∈ 0d−`+2 (x) ∩ 01 (v`−1 ) ∩ 01 (v`+1 ). Fix a path wd−`+2 = w, wd−`+3 , . . . , wd such that
wi ∈ 0i (x) (such a path exists since bi > 0 (0 ≤ i ≤ d − 1)). Suppose wd 6= v0 . Then
v0 , . . . , v`−1 , wd−`+2 , . . . , wd is a down-up path of length 2` − 2, contradicting the
minimality of length 2`. Thus wd = v0 . Similarly, v2` = wd , contradicting v0 6= v2` . It
2
follows that |0d (x)| = 1, so ad = 0. Hence 0 is bipartite in light of Lemma 4.6.
Lemma 4.9 0 is 2-homogeneous.
Proof: By [3, Theorem 16] it is enough to show that for all i (1 ≤ i ≤ d) and for all y,
z ∈ 0i (x) with ∂(y, z) = 2, the number γ (y, z) of (8) is independent of the choice of y, z.
Fix i (1 ≤ i ≤ d − 1), and pick any y, z ∈ 0i (x) with ∂(y, z) = 2. By Lemma 4.8,
0 is bipartite, so β(y, z) + γ (y, z) = c2 . By (17), si β(y, z) − si−1 γ (y, z) = 0. Thus
(si + si−1 )γ (y, z) = c2 si . Since si 6= 0 by Lemma 4.3, the right side is nonzero and hence
the left side is also nonzero. Thus we may solve this equation for γ (y, z) independent of y
and z. Observe that when i = d there is nothing to show by Lemma 4.8.
2
5.
Proof of Theorem 1.1
In this section we prove Theorems 1.1 and 3.6. We continue with the notation and assumptions of Theorem 4.1 throughout this section. We begin by considering the uniqueness of
the U (sl(2)) and Uq (sl(2)) structures.
+
Lemma 5.1 Set si = xi− xi+1
(0 ≤ i ≤ d − 1). Then the scalars si (0 ≤ i ≤ d − 1) and
z i (0 ≤ i ≤ d) are uniquely determined up to the same scalar multiple.
Proof: Observe that for all i (0 ≤ i ≤ d) and for all y ∈ 0i (x), β(y, y) = bi and
γ (y, y) = ci , where β(y, y) and γ (y, y) are as in (8). Thus applying (17) with y = z gives
s0 = z 0 b0−1 , si bi − si−1 ci = z i (1 ≤ i ≤ d − 1), sd−1 cd = −z d .
(18)
Applying the relation X − X + − X + X − = Z to K i (1 ≤ i ≤ d − 1) and simplifying with
Lemma 4.2 gives
si bi ci+1 − si−1 bi−1 ci = z i
(1 ≤ i ≤ d − 1).
(19)
Fix i (1 ≤ i ≤ d − 1). Subtracting (18) from (19) gives si bi (ci+1 − 1) = si−1 ci (bi−1 − 1).
Since si , si−1 are nonzero by Lemma 4.3, bi−1 = 1 if and only if ci+1 = 1. Suppose bi−1 =
ci+1 = 1. Then 1 ≤ bi ≤ bi−1 = 1 and 1 ≤ ci ≤ ci+1 = 1 since the ci form a nondecreasing
sequence and the bi form a nonincreasing sequence by [2, Proposition 4.1.6]. Thus k = ci +
bi = 2, a contradiction. Thus we may solve for si as si = ci (bi−1 − 1)si−1 /(bi (ci+1 − 1)).
In particular, since s0 = z 0 b0−1 , the numbers s j (0 ≤ j ≤ d − 1) are determined by the
intersection numbers and z 0 . The numbers z j (1 ≤ j ≤ d) are determined by (18). In
these formulas z 0 is a factor of s j (0 ≤ j ≤ d − 1) and z j (1 ≤ j ≤ d), so the result
follows.
2
34
CURTIN AND NOMURA
Lemma 5.2 Suppose that 0 P
is isomorphic to the d-cube. Then, after multiplying X + and
d
(d − 2i)E i∗ and X − , X + , Z satisfy (1).
Z by some same scalar, Z = i=0
Proof: By (16), si LREi∗ − si−1 RLEi∗ = z i E i∗ (0 ≤ i ≤ d), and by (10), LREi∗ − RLEi∗ =
(d − 2i)E i∗ (0 ≤ i ≤ d). One possibility is si = 1 (0 ≤ i ≤ d − 1), and in this case
z i = d − 2i (0 ≤ i ≤ d). Thus by Lemma 5.1, there exists a scalar α such that αsi = 1
− 2i (0 ≤ i ≤ d). Hence, after replacing X + with α X + and
(0 ≤ i ≤ d − 1) and αz i = d P
d
(d − 2i)E i∗ and X − , X + , and Z satisfy (1).
2
Z with α Z , we find that Z = i=0
Lemma 5.3 Suppose 0 is P
not isomorphic to the d-cube. Then, after
X + and Z
Pdmultiplying
d
∗
−
+
d−2i ∗
E i satisfy (4)
by some same scalar, Z = i=0 [ d − 2i ]E i and X , X , Y = i=0 q
for some real number q 6∈ {0, 1, −1}.
−1
LREi∗ −
Proof: By (16), si LREi∗ − si−1 RLEi∗ = z i E i∗ (0 ≤ i ≤ d), and by (15), ei−1 ei+1
−1 −1
∗
∗
ei−1 ei RLEi = [ d − 2i ]E i (0 ≤ i ≤ d), where e j and [ j ] are as in (12) for all
−1
(0 ≤ i ≤ d − 1), and in this case z i = [ d − 2i ]
integers j. One possibility is si = ei−1 ei+1
−1 −1
(0 ≤ i ≤ d). Thus by Lemma 5.1, there exists a scalar α such
Pdthat αsi = e∗i ei+1
(0 ≤ i ≤ d − 1) and αz i = [ d − 2i ] (0 ≤ i ≤ d). Hence α Z = i=0 [ d − 2i ]E i , and,
Pd
q d−2i E i∗ satisfy (4).
after replacing X + with α X + , we find that X − , X + , and Y = i=0
2
Lemma 5.4 The conclusions of Lemma 5.3 do not hold when 0 is isomorphic to the
d-cube, and the conclusions of Lemma 5.2 do not hold when 0 is not isomorphic to the
d-cube.
Proof: If this is not the case, then arguing as in Lemmas 5.2 and 5.3, we find that there
is a scalar α such that α(d − 2i) = [ d − 2i ] (0 ≤ i ≤ d), where [ d − 2i ] is as in (12)
for some real number q 6∈ {0, 1, −1}. When d is odd, this equation at i = (d − 1)/2
and i = (d − 3)/2 routinely implies q ∈ {1, −1}, and when d is even, this equation at
i = d/2 − 1 and i = d/2 − 2 routinely implies q ∈ {1, −1}, a contradiction.
2
We are ready to prove Theorems 1.1 and 3.6.
Proof of Theorem 3.6:
(i)⇒(ii): Observe that 0 is isomorphic to the d-cube by Theorem 4.1 and Lemma 5.4.
Applying the relation ZX − X − Z = 2X − to K i (1 ≤ i ≤ d) and simplifying with
−
−
−
bi−1 K i−1 − z i xi−1
bi−1 K i−1 = 2xi−1
bi−1 K i−1 (1 ≤ i ≤
Lemma 4.2, we find that z i−1 xi−1
d). Thus z i = z i−1 − 2 (1 ≤ i ≤ d), so z i = β + d − 2i (0 ≤ i ≤ d), where β = z 0 − d.
By Lemma 5.2, there exists a scalar α such that αz i = d − 2i (0 ≤ i ≤ d). Comparing
these formulas for z i , we find that α = 1 and β = 0. It follows from Lemma 5.2 that
z i = d − 2i (0 ≤ i ≤ d) and si = 1 (0 ≤ i ≤ d − 1).
(ii)⇒(i): The relations are verified exactly as in Lemma 3.2. We may argue as in Lemma
3.5 to show that these matrices generate T .
2
35
QUANTUM ENVELOPING ALGEBRA
Proof of Theorem 1.1:
(i)⇒(ii): 0 is bipartite and 2-homogeneous by Theorem 4.1. Note that it is not isomorphic
to the d-cube by assumption. We apply our results to U p (sl(2)) and use q to denote the
+
parameter of Theorem 3.3 while showing that the formulas for ( p + p −1 )2 and xi− xi+1
hold.
Applying the relation YX − = p 2 X − Y to K i (1 ≤ i ≤ d) and simplifying with Lemma
−
−
4.2, we find that yi−1 xi−1
bi−1 K i−1 = p 2 yi xi−1
bi−1 K i−1 (1 ≤ i ≤ d). Thus yi =
−2
d−2i
(0 ≤ i ≤ d), where β = y0 p −d . By Lemma 5.3,
yi−1 p (1 ≤ i ≤ d), so yi = βp
there exists a scalar α such that α(yi − yi−1 )( p − p −1 )−1 = (q d−2i − q −d+2i )(q − q −1 )−1
(0 ≤ i ≤ d). Combining these formulas,
α(βpd−2i − β −1 p −d+2i )( p − p −1 )−1
= (q d−2i − q −d+2i )(q − q −1 )−1
(0 ≤ i ≤ d).
(20)
Suppose d is odd. Then (20) at i = (d − 1)/2 and i = (d + 1)/2 routinely implies that
α = β ∈ {1, −1}. Now (20) at i = (d − 3)/2 gives p2 + p −2 = q 2 + q −2 . Suppose d is
even. Then (20) at i = d/2 routinely implies that β ∈ {1, −1}. Now (20) at i = d/2 − 1
and i = d/2 − 2 routinely implies that α = β and p2 + p −2 = q 2 + q −2 . In both cases
( p + p −1 )2 = (q + q −1 )2 , so the formula for ( p + p −1 )2 follows from Corollary 3.4.
+
follows from Lemma 5.3 (with ² = α).
The formula for xi− xi+1
+
−1
in (ii) equals ²ei−1 ei+1
(ii)⇒(i): Identical to Lemma 3.5 since the expression for xi− xi+1
2
(0 ≤ i ≤ d − 1).
6.
Remarks
The 2-homogeneous bipartite distance-regular graphs are essentially known.
Theorem 6.1 [8, 11] Let 0 = (X, R) denote distance-regular graph with diameter d ≥ 3
and valency k ≥ 3, and assume that 0 is not isomorphic to the d-cube. Then 0 is bipartite
and 2-homogeneous if and only if it is one of the following:
(i) the complement of the 2 × (k + 1)-grid;
(ii) a Hadamard graph of order 4γ for some positive integer γ ;
(iii) a bipartite distance-regular graph with diameter 5 and intersection array
{b0 , b1 , . . . , b4 ; c1 , c2 , . . . , c4 } = {k, k − 1, k − µ, µ, 1; 1, µ, k − µ, k − 1, k},
where k = γ (γ 2 + 3γ + 1), µ = γ (γ + 1) for some integer γ ≥ 2.
When γ = 2, (iii) is uniquely realized by the antipodal 2-cover of the Higman-Sims graph.
No examples of (iii) with γ ≥ 3 are known.
We present some examples of distance-regular graphs related to U (sl(2)) and Uq (sl(2))
which do not satisfy hypotheses of Theorem 1.1.
Let 0 = (X, R) denote the 2d-cycle (d ≥ 2). Fix x ∈ X , and write E i∗ = E i∗ (x) (0 ≤
i ≤ d) and T = T (x). Observe that 0 is vacuously 2-homogeneous. Let q be a primitive
36
CURTIN AND NOMURA
Pd−1
Pd
∗
∗
2dth root
and set X − = i=0
[d − i]E i∗ AE i+1
, X + = i=1
[i]E i∗ AE i−1
, and
Pd of unity,
d−2i ∗
−
+
Y = i=0 q
E i . Then X , X , and Y satisfy (4). However, these matrices do not
generate T . The 4-cycle is exceptional. In addition to the U (sl(2)) structure of Theorem
3.6, the 4-cycle has the Uq (sl(2)) structure of Theorem 1.1 for any non-zero complex
number q such that q 4 6= 1.
Let 0 = (X, R) denote the Hamming graph H (d, n), n ≥ 3. Fix x ∈ X , and write
E i∗ = E i∗ (x) (0 ≤ i ≤ d) and T = T (x). By [10, p. 202], X − = L, X + = R, and
Z = LR − RL satisfy (1). However, these matrices do not generate T and Z 6∈ M∗ .
It is hoped that some further light will be shed upon the Q-polynomial distance-regular
graphs through our work on the 2-homogeneous bipartite distance-regular graphs. Thus in
a future paper we will relate the algebraic properties of T to those of Uq (sl(2)).
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